Question: The lifespans of turtles in a particular zoo are normally distributed. The average turtle lives $91$ years; the standard deviation is $21.4$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a turtle living longer than $48.2$ years.
Answer: $91$ $69.6$ $112.4$ $48.2$ $133.8$ $26.8$ $155.2$ $95\%$ $2.5\%$ $2.5\%$ We know the lifespans are normally distributed with an average lifespan of $91$ years. We know the standard deviation is $21.4$ years, so one standard deviation below the mean is $69.6$ years and one standard deviation above the mean is $112.4$ years. Two standard deviations below the mean is $48.2$ years and two standard deviations above the mean is $133.8$ years. Three standard deviations below the mean is $26.8$ years and three standard deviations above the mean is $155.2$ years. We are interested in the probability of a turtle living longer than $48.2$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $95\%$ of the turtles will have lifespans within 2 standard deviations of the average lifespan. The remaining $5\%$ of the turtles will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({2.5\%})$ will live less than $48.2$ years and the other half $({2.5\%})$ will live longer than $133.8$ years. The probability of a particular turtle living longer than $48.2$ years is ${95\%} + {2.5\%}$, or $97.5\%$.